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I'm trying to prove $\exists x \exists y (S(x,y) \lor S(y,x)) \vdash \exists x \exists y S(x,y)$ in natural deduction, and I have already applied existential elimination to get $S(x_0,y_0)$, with $x_0$ and $y_0$ being the assumptions.

Yet I'm stuck on how to prove $S(x_0,y_0)$ from $S(x_0,y_0) \lor S(y_0,x_0)$ after getting $S(x_0,y_0) \lor S(y_0,x_0)$ from existential elimination.

Can someone help me out or is there other ways to approach this question?

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  • $\begingroup$ Are you using natural deduction? $\endgroup$ – Taroccoesbrocco Nov 5 '18 at 8:04
  • $\begingroup$ Because there are multiple formal proof systems, it is important to say exactly which system you are using. $\endgroup$ – Carl Mummert Nov 5 '18 at 12:26
  • $\begingroup$ @Taroccoesbrocco Yes. Sorry for not making it clear. $\endgroup$ – Ang. Nov 6 '18 at 8:41
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From $S(x_0,y_0)$ you conclude $\exists x \exists y S(x,y)$ by using existential generalization twice, once to introduce $\exists y$ and once more to introduce $\exists x$. From $S(y_0,x_0)$ you also conclude $\exists x\exists y S(x,y)$. Therefore you can conclude $\exists x\exists y S(x,y)$ from $S(x_0,y_0) \lor S(y_0,x_0)$.

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  • $\begingroup$ Do you mean by imagining doing S[$x_0$/y] to get S($y_0$, $x_0$) ? $\endgroup$ – Ang. Nov 5 '18 at 7:27
  • $\begingroup$ Yes, apply 'existential generalization' twice: once to introduce $\exists y$ and once more to introduce $\exists x$. $\endgroup$ – Magdiragdag Nov 5 '18 at 8:44
  • $\begingroup$ Okay, thank you so much for this! $\endgroup$ – Ang. Nov 5 '18 at 9:12
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Here is a proof using a Fitch-style proof checker:

enter image description here

Note that I used universal elimination twice from the same line 3. This allowed me to ultimately derive a contradiction on line 13 from the two disjuncts on line 9.


Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

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    $\begingroup$ Well, that is valid, but it is a rather roundabout route. You can do it directly with just assumptions, disjunction elimination, existential introductions, and existential eliminations. $\endgroup$ – Graham Kemp Aug 25 at 3:04
  • $\begingroup$ @GrahamKemp There may be alternate ways to do this. That this works is all I'm aiming for. $\endgroup$ – Frank Hubeny Aug 25 at 3:41

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